Currently I am reading Dembo's Probability Theory notes.
At page $9,$ the author provided exercise $1.1.8.$
Exercise $1.1.8$ Suppose $P$ is non-atomic and $A \in \mathcal{F}$ with $P(A) > 0.$
(a) Show that for every $\varepsilon>0$, we have $B\subseteq A$ such that $0<P(B)<\varepsilon$.
Definition $1.1.7$ We say that a probability space $(\Omega,\mathcal{F},P)$ is non-atomic, or alternatively call $P$ non-atomic if $P(A) > 0$ implies the existence of $B \in \mathcal{F}$, $B \subseteq A$ with $$0 < P(B) < P(A).$$
My attempt:
By atomicity of $P,$ there exists $A_1\subseteq A$ such that $$0<P(A_1)<P(A).$$ Express $$A = A_1 \cup (A\setminus A_1).$$ If either $A_1$ or $A\setminus A_1$ have measures less than $\varepsilon,$ then we are done. Otherwise, assume that $P(A_1)\geq \varepsilon$ and $P(A\setminus A_1)\geq \varepsilon.$
By non-atomicity of $P$ on $A\setminus A_1,$ there exists $A_2\subseteq A \setminus A_1$ such that $0<P(A_2)<P(A\setminus A_1)$ and $$A\setminus A_1 = A_2 \cup [(A\setminus A_1)\setminus A_2].$$ If either $P(A_2)<\varepsilon$ or $P[(A\setminus A_1)\setminus A_2]< \varepsilon,$ then we are done. Assume otherwise.
If at some point we have $A_N\subseteq A$ such that $P(A_N)<\varepsilon,$ then we are done.
Otherwise, we obtain a sequence of subsets $(A_n)_{n=1}^\infty$ of $A$ such that $$A = \bigcup_{n=1}^\infty A_n$$ where the union is disjoint and $P(A_n)\geq \varepsilon$ for all $n\in\mathbb{N}.$ However, by countably additivity of $P,$ we have $$P(A) = \sum_{n=1}^\infty P(A_n) \geq \sum_{n=1}^\infty \varepsilon = \infty,$$ a contradiction. Therefore, the process must terminate and hence there exists $N\in\mathbb{N}$ such that $P(A_N)<\varepsilon.$
Is my attempt correct?
You could probably avoid using Reductio ad absurdum by noticing that $P(A_N) < P(A)-\varepsilon N$ thus your algorithm stops after at most $\left[\frac{P(A)}{\varepsilon}\right]+1$ steps.